2 Answers
2

You want to start with $(x-2)(x-3)(x+5)$ to make sure the zeroes are correct. Your extra degree of freedom comes from the fact that $c(x-2)(x-3)(x+5)$ also has those roots for any constant $c$.

If the function is to pass through $(4,36)$, then you want to solve
$$
c(4-2)(4-3)(4+5) = 36
$$
for $c$. I get $c = 2$.

Your function is entirely specified now: $f(x) = 2(x-2)(x-3)(x+5)$.

To find out where it has a value of 120, you want to solve
$$
120 = 2(x-2)(x-3)(x+5)
$$
for $x$. Depending on what options you have at your disposal, you can do this either by graphing or by setting the function equal to 0 and using a combination of rational roots tests, polynomial division, and factoring.

yeah i got the answer except one minor difference the c value would be 2 since 36 = a(4-2)(4-1)(4+5) 36 =18a a =2 for the rest of the roots i just solved for when both sides =0 after moving 120/2 over then syntehtic divison (a, c i assume they both mean the same)
–
FarazOct 12 '11 at 5:38

I get $c = 2$ as well. I flubbed up some numbers on my first try.
–
Austin MohrOct 12 '11 at 5:41